Passing an arbitrary object to Coffeescript's anonymous function

Question

When I run the following Coffeescript code:

@sum = (x, y) -> x + y

I get this compiled Javascript:

(function() {

    this.sum = function(x, y) {
        return x + y;
    };

}).call(this);

Is there a way in Coffeescript to replace this in .call(this) with an arbitrary object like myObjector anything?

Your code is incomplete. Please add what's before/after @sum = (x, y) -> x + y1 — Robin Maben
– Robin Maben, Commented Aug 24, 2012 at 9:23
No, this is actually my complete code. This is why I am asking how to change the this object, so that sum() is added to whatever object I want. — Amberlamps
– Amberlamps, Commented Aug 24, 2012 at 9:25
(function() { and }).call(this); are not compiled from @sum = .... These are the result of running coffee without the --bare flag. this is the actual result from compiling. — Rob W
– Rob W, Commented Aug 24, 2012 at 9:30
@RobW: I did not know that. I run coffee -c example.coffee in my console and I always get that anonymous function wrapper. I thought that was standard. — Amberlamps
– Amberlamps, Commented Aug 24, 2012 at 9:32

Rob W · Accepted Answer · 2012-08-24 09:39:38Z

1

(function() { and }).call(this); are not the result of compiling @sum = ..., but added by the coffee executable. This is the actual result from compiling:

this.sum = function(x, y) {
  return x + y;
};

To get a different/desired output, run coffee -b -c (or coffee -bc or coffee --bare --compile) using the following code:

(-> 
  @sum = (x, y) -> x + y
).call WHATEVER

becomes

(function() {
  return this.sum = function(x, y) {
    return x + y;
  };
}).call(WHATEVER);

edited Aug 24, 2012 at 9:39

answered Aug 24, 2012 at 9:34

Rob W

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1 Comment

Amberlamps Over a year ago

That is pretty much what I was looking for. Thanks! The flag does the trick.

Robin Maben · Accepted Answer · 2012-08-24 09:33:23Z

1

myobj.sum = (x, y) -> x + y

should get compiled to (UPDATE: See Rob W's answer for compile options) :-

myobj.sum = function(x, y) {
  return x + y;
};

Isn't that what you want? So further you can call it using myobj.sum a, b

Complete code..

myobj = {}
myobj.sum = (x, y) -> x + y

alert(myobj.sum 10,4)

edited Aug 24, 2012 at 9:33

answered Aug 24, 2012 at 9:26

Robin Maben

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2 Comments

Amberlamps Over a year ago

No, I want to know whether that anonymous function can in any way be modified like for instance passing a different object than this

Robin Maben Over a year ago

@Amberlamps: Why, is it because you want to extend the sum function into whatever object you pass? And then again, that anonymous function cannot ever be called again, so I don't see the point.

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